Thanks for your answer. Again, I'm not sure if I understood this correctly.
I used c as I thought s = c for simple ions. I now took the root of √I to get I.
I assumed the C for KNO3 = 0.0050 and C for CaSO4 = s = 0.0154
This might be where I am wrong?
I don't really know how to begin. I've done alkylations by having two of the same compounds react with each other e.g. two aldehydes but never started out with dimethyl malonate.
I was thinking I need 1,4 dibromobutane to form the cyclopentane ring but apart from that I'm clueless
Apologies, that makes sense! It seems like I put the equation the wrong way around ie calculated 24.305/60.8 instead of 60.8/24.305.
So the moles for each of the compounds are 2.5, leaving me with 453.28g of phenyl magnesium bromide
The equivalent numbers were probably this wrong because I...
I have worked the following out: 72.5g of phenyl magnesium bromide are prepared from the reagents.
I tried to take it a step further and create the same table for the reaction of phenyl magnesium bromide and 2-phenyl propanal but I'm unsure whether I can infer it like that - especially given...
Am I right to assume that the equivalents for phenyl magnesium bromide would be 3 (magnesium, bromobenzene and ether (?)) And the number for 2-phenyl-propanal 1 as it says "...was added to 2-phenylpropionaldehyde" which is the same thing as 2-phenylpropanal? I don't know if I understood this...
Oh you are right, the C would be lower in energy, therefore under the SEt orbital.
And since the bond with sodium breaks, it seems like it only has an unpaired electron left
I think I've come a bit farther this time. The bonding/anti bonding hint really helped! :)
Two questions: is there only one or two e- in the orbital of SEt- ? It is a lone pair, so there would be 2 e-, correct?
And would the orbital diagram drawn enough to answer the question combined with...
Thanks for your input!
I know it's not a lot but I started working upon your prompts. In the first one is the orbital attacking the C centre with the positive charge after Br has left. Maybe it would be good to but Br- on the side too?
The second one is my attempt at the orbital diagram...
The equation can be seen in the picture. I was assuming it was an Sn2 reaction given the polar aprotic solvent and strong nucleophile.
For b), I assumed that the question asked me to draw the orbitals as shown above (as opposed to orbital diagram) because it asked me to identify the HOMO and...